0
$\begingroup$

enter image description here

Can someone explain how to solve this problem. Since I can get 7 integers from 1st inequality 5 integers from 2nd inequality I got total number of cases of 35. Then I counted the possibilities that would be less than 4. For example x=o y can = 0, 1, 2, 3, when x=1 y=0,1,2 and so on. I came down to 10/35, but the answer is 1/3 .

$\endgroup$
6
  • $\begingroup$ This problem isn't very clear. Are $x$ and $y$ supposed to be random variables? If so, what are their distributions? $\endgroup$ – Math1000 Dec 6 '14 at 2:34
  • $\begingroup$ Sorry, that was all I was given. I believe they don't have any distribution and that they're integers. $\endgroup$ – GMATnoob Dec 6 '14 at 2:37
  • $\begingroup$ Can you first confirm that $x$ and $y$ are integers, not just real numbers? This makes a difference. $\endgroup$ – peterwhy Dec 6 '14 at 2:42
  • 3
    $\begingroup$ The answer is $\frac{1}{3}$ if they're real numbers with uniform distribution. (A four by six rectangle has area 24, the right triangle cut off by $x+y < 4$ and $x,y \geq 0$, with side lengths four and four, has area 8.) $\endgroup$ – aes Dec 6 '14 at 2:42
  • $\begingroup$ Confirmed they are real numbers with uniform dist. aes, could you explain further? not sure how you got side lengths four and four. $\endgroup$ – GMATnoob Dec 6 '14 at 3:03
2
$\begingroup$

Here's a six by four rectangle, representing the possible values of $x$ and $y$. Choosing $x$ and $y$ uniformly and independently means choosing a random point in this rectangle, with the probability it's in a given region proportional to the area of that region.

enter image description here

I've drawn a line $x+y = 4$. The values you're interested in are below that line.

The total area is 24, and the area of the triangle below $x+y=4$ is $8$.

Thus the probability that $x+y < 4$ is $\frac{8}{24} = \frac{1}{3}$.

$\endgroup$
2
  • $\begingroup$ x+y=4 y=-x+4 so 4=y-intercept and x has slope of -1/1. Makes perfect sense. The answer with the visual was very helpful. Thanks a bunch. $\endgroup$ – GMATnoob Dec 6 '14 at 3:32
  • $\begingroup$ @GMATnoob Certainly. Also, you could look at $x+y=4$ differently: both $(4,0)$ and $(0,4)$ are solutions, and it's linear, so the line is the line through those two points. $\endgroup$ – aes Dec 6 '14 at 3:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.